Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

<image>

You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
template <typename T>
inline T max(T a, T b) {
  return a > b ? a : b;
}
template <typename T>
inline T min(T a, T b) {
  return a < b ? a : b;
}
template <typename T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
template <typename T>
inline bool repr(T &a, T b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline bool repl(T &a, T b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline T gcd(T a, T b) {
  T t;
  if (a < b) {
    while (a) {
      t = a;
      a = b % a;
      b = t;
    }
    return b;
  } else {
    while (b) {
      t = b;
      b = a % b;
      a = t;
    }
    return a;
  }
}
template <typename T>
inline T sqr(T x) {
  return x * x;
}
struct Cg {
  inline char operator()() { return getchar(); }
};
struct Cp {
  inline void operator()(char x) { putchar(x); }
};
inline bool IS(char x) { return x == 10 || x == 13 || x == ' '; }
template <typename T>
struct Fr {
  T P;
  inline Fr &operator,(int &x) {
    x = 0;
    unsigned char t = P();
    while ((t < '0' || t > '9') && t != '-') t = P();
    bool f = 0;
    if (t == '-') t = P(), f = 1;
    x = t - '0';
    for (t = P(); t >= '0' && t <= '9'; t = P()) x = x * 10 + t - '0';
    if (f) x = -x;
    return *this;
  }
  inline operator int() {
    int x;
    *this, x;
    return x;
  }
  inline Fr &operator,(long long &x) {
    x = 0;
    unsigned char t = P();
    while ((t < '0' || t > '9') && t != '-') t = P();
    bool f = 0;
    if (t == '-') t = P(), f = 1;
    x = t - '0';
    for (t = P(); t >= '0' && t <= '9'; t = P()) x = x * 10 + t - '0';
    if (f) x = -x;
    return *this;
  }
  inline operator long long() {
    long long x;
    *this, x;
    return x;
  }
  inline Fr &operator,(char &x) {
    for (x = P(); IS(x); x = P())
      ;
    return *this;
  }
  inline operator char() {
    char x;
    *this, x;
    return x;
  }
  inline Fr &operator,(char *x) {
    char t = P();
    for (; IS(t); t = P())
      ;
    if (~t) {
      for (; !IS(t) && ~t; t = P()) *x++ = t;
    }
    *x++ = 0;
    return *this;
  }
  inline Fr &operator,(double &x) {
    x = 0;
    unsigned char t = P();
    while ((t < '0' || t > '9') && t != '-') t = P();
    bool f = 0;
    if (t == '-') t = P(), f = 1;
    x = t - '0';
    for (t = P(); t >= '0' && t <= '9'; t = P()) x = x * 10 + t - '0';
    if (t == '.') {
      double u = 0.1;
      for (t = P(); t >= '0' && t <= '9'; t = P(), u *= 0.1) x += u * (t - '0');
    }
    if (f) x = -x;
    return *this;
  }
  inline operator double() {
    double x;
    *this, x;
    return x;
  }
  inline Fr &operator,(long double &x) {
    x = 0;
    unsigned char t = P();
    while ((t < '0' || t > '9') && t != '-') t = P();
    bool f = 0;
    if (t == '-') t = P(), f = 1;
    x = t - '0';
    for (t = P(); t >= '0' && t <= '9'; t = P()) x = x * 10 + t - '0';
    if (t == '.') {
      double u = 0.1;
      for (t = P(); t >= '0' && t <= '9'; t = P(), u *= 0.1) x += u * (t - '0');
    }
    if (f) x = -x;
    return *this;
  }
  inline operator long double() {
    long double x;
    *this, x;
    return x;
  }
  inline Fr &operator,(unsigned int &x) {
    x = 0;
    unsigned char t = P();
    while (t < '0' || t > '9') t = P();
    x = t - '0';
    for (t = P(); t >= '0' && t <= '9'; t = P()) x = x * 10 + t - '0';
    return *this;
  }
  inline operator unsigned int() {
    unsigned int x;
    *this, x;
    return x;
  }
  inline Fr &operator,(unsigned long long &x) {
    x = 0;
    unsigned char t = P();
    while (t < '0' || t > '9') t = P();
    x = t - '0';
    for (t = P(); t >= '0' && t <= '9'; t = P()) x = x * 10 + t - '0';
    return *this;
  }
  inline operator unsigned long long() {
    unsigned long long x;
    *this, x;
    return x;
  }
};
Fr<Cg> in;
template <typename T>
struct Fw {
  T P;
  inline Fw &operator,(int x) {
    if (x) {
      if (x < 0) P('-'), x = -x;
      unsigned char s[10], c = 0;
      while (x) s[c++] = x % 10 + '0', x /= 10;
      while (c--) P(s[c]);
    } else
      P('0');
    return *this;
  }
  inline Fw &operator()(int x) {
    if (x) {
      if (x < 0) P('-'), x = -x;
      unsigned char s[10], c = 0;
      while (x) s[c++] = x % 10 + '0', x /= 10;
      while (c--) P(s[c]);
    } else
      P('0');
    return *this;
  }
  inline Fw &operator,(unsigned int x) {
    if (x) {
      unsigned char s[10], c = 0;
      while (x) s[c++] = x % 10 + '0', x /= 10;
      while (c--) P(s[c]);
    } else
      P('0');
    return *this;
  }
  inline Fw &operator()(unsigned int x) {
    if (x) {
      unsigned char s[10], c = 0;
      while (x) s[c++] = x % 10 + '0', x /= 10;
      while (c--) P(s[c]);
    } else
      P('0');
    return *this;
  }
  inline Fw &operator,(long long x) {
    if (x) {
      if (x < 0) P('-'), x = -x;
      unsigned char s[19], c = 0;
      while (x) s[c++] = x % 10 + '0', x /= 10;
      while (c--) P(s[c]);
    } else
      P('0');
    return *this;
  }
  inline Fw &operator()(long long x) {
    if (x) {
      if (x < 0) P('-'), x = -x;
      unsigned char s[19], c = 0;
      while (x) s[c++] = x % 10 + '0', x /= 10;
      while (c--) P(s[c]);
    } else
      P('0');
    return *this;
  }
  inline Fw &operator,(unsigned long long x) {
    if (x) {
      unsigned char s[20], c = 0;
      while (x) s[c++] = x % 10 + '0', x /= 10;
      while (c--) P(s[c]);
    } else
      P('0');
    return *this;
  }
  inline Fw &operator()(unsigned long long x) {
    if (x) {
      unsigned char s[20], c = 0;
      while (x) s[c++] = x % 10 + '0', x /= 10;
      while (c--) P(s[c]);
    } else
      P('0');
    return *this;
  }
  inline Fw &operator,(char x) {
    P(x);
    return *this;
  }
  inline Fw &operator()(char x) {
    P(x);
    return *this;
  }
  inline Fw &operator,(const char *x) {
    while (*x) P(*x++);
    return *this;
  }
  inline Fw &operator()(const char *x) {
    while (*x) P(*x++);
    return *this;
  }
  inline Fw &operator()(double x, int y) {
    if (y) {
      double t = 0.5;
      for (int i = y; i--;) t *= 0.1;
      if (x >= 0)
        x += t;
      else
        x -= t, P('-');
      *this, (long long)(abs(x));
      P('.');
      if (x < 0) x = -x;
      while (y--) {
        x *= 10;
        x -= floor(x * 0.1) * 10;
        P(((int)x) % 10 + '0');
      }
    } else if (x >= 0)
      *this, (long long)(x + 0.5);
    else
      *this, (long long)(x - 0.5);
    ;
    return *this;
  }
  inline Fw &operator()(long double x, int y) {
    if (y) {
      double t = 0.5;
      for (int i = y; i--;) t *= 0.1;
      if (x >= 0)
        x += t;
      else
        x -= t, P('-');
      *this, (long long)(abs(x));
      P('.');
      if (x < 0) x = -x;
      while (y--) {
        x *= 10;
        x -= floor(x * 0.1) * 10;
        P(((int)x) % 10 + '0');
      }
    } else if (x >= 0)
      *this, (long long)(x + 0.5);
    else
      *this, (long long)(x - 0.5);
    ;
    return *this;
  }
};
Fw<Cp> out;
const int N = 400007;
int n, t, a, b, s_[N], p[N];
double s[N];
int main() {
  in, t, a, b, n;
  for (int i = 1, iend = n; i <= iend; i++) in, s_[i * 2 - 1], s_[i * 2];
  for (int i = 1, iend = n; i <= iend; i++)
    p[i] = p[i - 1] + s_[i * 2] - s_[i * 2 - 1];
  for (int i = 1, iend = n * 2; i <= iend; i++) s[i] = s_[i];
  for (int m = in; m--;) {
    int x, y;
    in, x, y;
    double l = x + (double)y / (y - t) * (a - x),
           r = x + (double)y / (y - t) * (b - x);
    int tl = std::lower_bound(s + 1, s + n * 2 + 1, l) - s,
        tr = std::lower_bound(s + 1, s + n * 2 + 1, r) - s;
    if (tl == n * 2 + 1 || tr == 1) {
      out, 0, '\n';
      continue;
    }
    if (tl == tr) {
      if (tl & 1)
        out, 0, '\n';
      else
        out, b - a, '\n';
      continue;
    }
    tr--;
    double ans = 0;
    if (~tl & 1) {
      ans += s[tl] - l;
      tl++;
    }
    if (tr & 1) {
      ans += r - s[tr];
      tr--;
    }
    ans += p[tr / 2] - p[(tl - 1) / 2];
    printf("%.10lf\n", ans / (r - l) * (b - a));
  }
}